9
$\begingroup$

Let $F$ be a non-abelian free group, that is, a free group of rank at least $2$, and let $\phi: F \rightarrow \mathbb{Z}$ be a nontrivial group homomorphism. How to prove that the kernel of $\phi$ is not finitely generated?

$\endgroup$

3 Answers 3

4
$\begingroup$

There is a useful result you could use to prove this.

Let $ F $ be a free group and $ E \leq F $ with $ |F:E| = \infty $. Suppose that $ \exists \: \{1\} \neq N \unlhd G $ with $ N \leq E $. Then the rank of $E$ is infinite.

The proof is as follows and assumes familiarity with the theory of Schreier generators of subgroups of free groups.

Let $F$ be free on $X$, let $ U $ be a Schreier transversal of $ E $ in $ F $ and, for $g \in F$, denote the element in $U \cap Eg$ by $\overline{g}$.

Let $ 1 \neq w = a_1 \cdots a_l \in N \leq E $, with $a_i \in X^{\pm 1}$. For $ u \in U $, $ Euw = Euwu^{-1}u = Eu $, since $ uwu^{-1} \in N \leq E $. So $ \overline{uw} = u $, and $ uw \not\in U $, so there is a least $ k $ such that $ ua_1 \cdots a_k \notin U $. Let $ u_k := ua_1 \cdots a_{k-1} $. Then $ u_k \in U $ and $ u_ka_k \notin U $, so $ u_ka_k\overline{u_ka_k}^{-1} $ is not trivial. Since $ U $ is infinite and $l$ is fixed, there is an infinite subset $ V $ of $ U $ and a fixed $k$ with $1 \le k \le l$, such that $k$ is minimal with $u_ka_k \notin U$ for all $u \in V$. Then $ \left\{ u_ka_k\overline{u_ka_k}^{-1} : u \in V \right\} $ is an infinite subset of the set of Schreier generators of $E$, and hence $E$ has infinite rank.

$\endgroup$
4
$\begingroup$

For a completely different proof, that uses only the universal property of free groups, you could factor your map $\phi:F \to {\mathbb Z}$ as $\phi:F \to H \to {\mathbb Z}$, where $H$ is the wreath product ${\mathbb Z} \wr {\mathbb Z}$, and observe that the kernel of $H \to {\mathbb Z}$ is infinitely generated.

$\endgroup$
4
  • $\begingroup$ I know this was a question from long ago, but how does the map $F \to \mathbb{Z}$ factor into $F \to H \to \mathbb{Z}?$ $\endgroup$ Apr 6, 2018 at 14:30
  • 1
    $\begingroup$ Write $H = \langle a,b \rangle$ with $b$ in the base group of the wreath product, and define the map $H \to {\mathbb Z}$ by $a \mapsto 1$, $b \mapsto 0$. Now find a free basis $x_1,x_2,\ldots$ of $F$ such that $\phi(x_1) = 1$ and $\phi(x_i)=0$ for $i>1$, and map $F \to H$ by $x_1 \mapsto a$, $x_i \mapsto b$ for $i>1$. $\endgroup$
    – Derek Holt
    Apr 6, 2018 at 15:06
  • $\begingroup$ You seem to be implying that any such $\phi$ is surjective since it hits $1$. Is this a general fact? $\endgroup$
    – Jason V
    May 20, 2022 at 11:27
  • $\begingroup$ Ah, I see what you mean. But for any nontrivial map $\phi: F \to {\mathbb Z}$, the image is isomorphic to ${\mathbb Z}$, so I can assume that $\phi$ is surjective. $\endgroup$
    – Derek Holt
    May 20, 2022 at 12:40
3
$\begingroup$

This follows since the rank of any infinite-index subgroup of a nonabelian free group is infinite. One way to prove this is by covering space theory. An infinite index subgroup corresponds to a cover of a wedge of circles where every point has infinitely many preimages. Then, if the cover is not a tree, there is at least one loop, $L$. Acting on $L$ by the infinite group of deck transformations gives infinitely many copies of $L$, but some of them may intersect or even coincide. However, since each loop has finitely many edges, only finitely many copies of $L$ can intersect $L$. So we can produce infinitely many disjoint loops. Thus the graph has infinite rank.

$\endgroup$
1
  • 1
    $\begingroup$ I think there might be a simpler proof, using elementary concepts. Anyway, thanks Jim Conant for the interesting answer. I am thinking about it. $\endgroup$
    – rla
    Apr 19, 2012 at 13:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .